Question

我正在通过做一些练习来学习可变参数模板，当涉及到lambda中的参数包扩展时，我感到困惑

所以，我的想法是编写一个有效负载可调用的计时器类，但是当我尝试在lambda函数中扩展参数包时出现编译错误。

gcc版本4.8.4（Ubuntu 4.8.4-2ubuntu1~14.04.3）

   template<typename F, typename... Args>
   struct timer
   {   
      timer(const std::chrono::milliseconds milliseconds, F call, Args&&... args)
      {   
         m_timer = std::make_shared<boost::asio::steady_timer>(
               timer_manager::instance().get_io_service(),
               std::chrono::steady_clock::now() + milliseconds
         );  

         //m_timer->async_wait(call, std::forward<Args>(args)...);
         m_timer->async_wait([=](const boost::system::error_code& ec){
             call(std::forward<Args>(args)...); //Error here
         }); 
      }   

      std::shared_ptr<boost::asio::steady_timer> m_timer;
   };


   // Helper to create the timer
   template<typename F, typename... Args>
   timer<F,Args...> create_timer(const std::chrono::milliseconds milliseconds,F callable, Args&& ...args)
   {   
      return timer<F,Args...>(milliseconds, std::forward<F>(callable), std::forward<Args>(args)...);
   }

使用它的主程序：

auto timer = timer::create_timer(std::chrono::milliseconds(5000), []()                                                                                                                  
{
   std::cout << "timer fired in main" << std::endl;
   payload::execute(10);           
 });

错误：

home/samdaniel/timer/src/timer.hpp: In lambda function:
/home/samdaniel/timer/src/timer.hpp:36:43: error: parameter packs not expanded with ‘...’:
                   call(std::forward<Args>(args)...);
                                           ^
/home/samdaniel/timer/src/timer.hpp:36:43: note:         ‘args’
/home/samdaniel/timer/src/timer.hpp: In instantiation of ‘struct timer::timer<F, Args>::timer(std::chrono::milliseconds, F, Args&& ...) [with F = main()::__lambda1; Args = {}; std::chrono::milliseconds = std::chrono::duration<long int, std::ratio<1l, 1000l> >]::__lambda0’:
/home/samdaniel/timer/src/timer.hpp:35:10:   required from ‘timer::timer<F, Args>::timer(std::chrono::milliseconds, F, Args&& ...) [with F = main()::__lambda1; Args = {}; std::chrono::milliseconds = std::chrono::duration<long int, std::ratio<1l, 1000l> >]’
/home/samdaniel/timer/src/timer.hpp:50:99:   required from ‘timer::timer<F, Args ...> timer::create_timer(std::chrono::milliseconds, F, Args&& ...) [with F = main()::__lambda1; Args = {}; std::chrono::milliseconds = std::chrono::duration<long int, std::ratio<1l, 1000l> >]’
/home/samdaniel/src/main.cpp:21:11:   required from here
/home/samdaniel/timer/src/timer.hpp:36:43: error: using invalid field ‘timer::timer<F, Args>::timer(std::chrono::milliseconds, F, Args&& ...)::__lambda0::__args’
make[2]: *** [src/CMakeFiles/coding_with_me.dir/main.cpp.o] Error 1
make[1]: *** [src/CMakeFiles/coding_with_me.dir/all] Error 2
make: *** [all] Error 2

新例子：

#include <iostream>
#include <functional>

namespace test
{
   template<typename F, typename... Args>
   void callback_dispatcher(F call, Args&& ...args )
   {   
      std::cout << "callback_dispatcher>" << __PRETTY_FUNCTION__ << "enter <<< " << std::endl;                                                                                                 
      auto invoke_me = [=](){
         call(std::forward<Args>(args)...);
      };  
      invoke_me();
   }   
}

int main()
{
   std::cout << "Main entered..." << std::endl;


   test::callback_dispatcher([](int a)
         {
         std::cout << "void(int) lambda dispatched with a = " << a << std::endl;
         },5);

   std::cout << "Main exited..." << std::endl;
}

错误：

src/generic_callback.cc: In lambda function:
src/generic_callback.cc:11:34: error: parameter packs not expanded with ‘...’:
          call(std::forward<Args>(args)...);
                                  ^
src/generic_callback.cc:11:34: note:         ‘args’
src/generic_callback.cc: In instantiation of ‘struct test::callback_dispatcher(F, Args&& ...) [with F = std::_Bind<std::_Mem_fn<void (plo_callback_tester::*)()>(std::_Placeholder<1>)>; Args = {plo_callback_tester* const}]::__lambda0’:
src/generic_callback.cc:12:7:   required from ‘void test::callback_dispatcher(F, Args&& ...) [with F = std::_Bind<std::_Mem_fn<void (plo_callback_tester::*)()>(std::_Placeholder<1>)>; Args = {plo_callback_tester* const}]’
src/generic_callback.cc:25:101:   required from here
src/generic_callback.cc:11:34: error: using invalid field ‘test::callback_dispatcher(F, Args&& ...)::__lambda0::__args’
src/generic_callback.cc: In instantiation of ‘struct test::callback_dispatcher(F, Args&& ...) [with F = main()::__lambda1; Args = {int}]::__lambda0’:
src/generic_callback.cc:12:7:   required from ‘void test::callback_dispatcher(F, Args&& ...) [with F = main()::__lambda1; Args = {int}]’
src/generic_callback.cc:41:13:   required from here
src/generic_callback.cc:11:34: error: using invalid field ‘test::callback_dispatcher(F, Args&& ...)::__lambda0::__args’

Answer 1

#include <iostream>

namespace test
{
  template<typename F, typename... Args>
  void callback_dispatcher(F call, Args&& ...args )
  {   
    std::cout << "callback_dispatcher>" << __PRETTY_FUNCTION__ << "enter <<< " << std::endl;                                                 
    auto invoke_me = [=]()mutable{
      call(std::move(args)...);
    };  
    invoke_me();
  }   
}

int main()
{
  std::cout << "Main entered..." << std::endl;


  test::callback_dispatcher(
    [](int a)
    {
      std::cout << "void(int) lambda dispatched with a = " << a << std::endl;
    },
    5
  );

  std::cout << "Main exited..." << std::endl;
}

上述微小变化在gcc 4.9及以上编译，但在gcc 4.8.5中没有。

它还会编译现代版本的clang。

更新编译器，或者不要使用编译器不支持的c ++ 11功能。

你可以将你的args打包成一个元组，然后将它们解压缩到函数体内。

  template<typename F, typename... Args>
  void callback_dispatcher(F call, Args&& ...args )
  {   
    std::cout << "callback_dispatcher>" << __PRETTY_FUNCTION__ << "enter <<< " << std::endl;                                                 
    auto invoke_me = [tup = std::make_tuple(std::forward<Args>(args)...), call=std::move(call)]()mutable{
      std::apply( call, std::move(tup) );
    };  
    invoke_me();
  }

您现在必须在C ++ 11的有限子集中实现std::apply，这是一个C ++ 17特性。（不要称之为std::apply）。这种方法可能适用于g ++ 4.8。

namespace notstd {
  template<std::size_t...Is> struct index_sequence {using type=index_sequence;};
  template<std::size_t N, std::size_t...Is> struct make_index_sequence:make_index_sequence<N-1, N-1, Is...>{};
  template<std::size_t...Is> struct make_index_sequence<0,Is...>:index_sequence<Is...>{};
  namespace details {
    template<class F, class Tuple, std::size_t...Is>
    auto apply( ::notstd::index_sequence<Is...>, F&& f, Tuple&& tuple )
    -> decltype( std::declval<F>()( std::get<Is>(std::declval<Tuple>())... ) )
    {
      return std::forward<F>(f)( std::get<Is>(std::forward<Tuple>(tuple))... );
    }
  }
  template<class F, class Tuple>
  auto apply( F&& f, Tuple&& tuple )
  -> decltype(::notstd::details::apply( make_index_sequence<std::tuple_size<typename std::decay<Tuple>::type>::value>{}, std::forward<F>(f), std::forward<Tuple>(tuple) ))
  {
    return ::notstd::details::apply( make_index_sequence<std::tuple_size<typename std::decay<Tuple>::type>::value>{}, std::forward<F>(f), std::forward<Tuple>(tuple) );
  }
}

live example of C++11 compliant notstd::apply

Answer 2

如果扩展在lambda中不起作用，则可以在创建功能对象期间将包扩展到外部，然后将其传递给该lambda。

所以代替：

m_timer->async_wait([=](const boost::system::error_code& ec){
    call(std::forward<Args>(args)...); //Error here
});

可能是：

auto functor = std::bind(std::forward<F>(call), std::forward<Args>(args)...);
m_timer->async_wait([=](const boost::system::error_code& ec){
    functor();
});

Answer 3

我在使用c ++ 11的c ++（GCC）4.8.5 20150623（Red Hat 4.8.5-36）中也遇到了相同的错误

template<typename... Args>
    void emplace_front(Args&&... args)
    {
        target( [&] {
                     _collection.emplace_front(std::forward<Args>(args)...);
                     } );
    }

以下为我工作！

template<typename... Args>
    void emplace_front(Args&&... args)
    {
        target( [&] (Args ...ar) -> void {
                           _collection.emplace_front(ar...);
                           } );
    }

我收到Args && ... args未使用但很容易解决的警告。

参数包扩展无法在lambda中工作

3 个答案: